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C++ copy constructor called at return

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C++ copy constructor called at return


When to use virtual destructors?In which situations is the C++ copy constructor called?What are the differences between a pointer variable and a reference variable in C++?Why don't C++ compilers define operator== and operator!=?What is the lifetime of a static variable in a C++ function?Can I call a constructor from another constructor (do constructor chaining) in C++?Inheriting constructorsHow can I profile C++ code running on Linux?The Definitive C++ Book Guide and ListWhen to use virtual destructors?What is the “-->” operator in C++?What is the copy-and-swap idiom?













9















error: use of deleted function 'A::A(const A&)'
 return tmp;
 ^~~


Why is the copy constructor called only when there is a virtual destructor in A? How to avoid this?



struct B ;

struct A
 std::unique_ptr<B> x;
 virtual ~A() = default;
;

A f() 
 A tmp;
 return tmp;






c++







share|improve this question









New contributor




Sobuch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • see: In which situations is the C++ copy constructor called?

    – kmdreko
    3 hours ago






  • 3





    C++ handles objects different than C#/Java. When an instance goes out of scope (tmp here) its destructor must be called. Therefore, when you return tmp then you're asking it to make a copy of tmp to be return to whomever calls the function. Once copied, tmp will be destroyed and its copy will be available for use.

    – Everyone
    3 hours ago















9















error: use of deleted function 'A::A(const A&)'
 return tmp;
 ^~~


Why is the copy constructor called only when there is a virtual destructor in A? How to avoid this?



struct B ;

struct A
 std::unique_ptr<B> x;
 virtual ~A() = default;
;

A f() 
 A tmp;
 return tmp;






c++







share|improve this question









New contributor




Sobuch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • see: In which situations is the C++ copy constructor called?

    – kmdreko
    3 hours ago






  • 3





    C++ handles objects different than C#/Java. When an instance goes out of scope (tmp here) its destructor must be called. Therefore, when you return tmp then you're asking it to make a copy of tmp to be return to whomever calls the function. Once copied, tmp will be destroyed and its copy will be available for use.

    – Everyone
    3 hours ago













9












9








9


1






error: use of deleted function 'A::A(const A&)'
 return tmp;
 ^~~


Why is the copy constructor called only when there is a virtual destructor in A? How to avoid this?



struct B ;

struct A
 std::unique_ptr<B> x;
 virtual ~A() = default;
;

A f() 
 A tmp;
 return tmp;






c++







share|improve this question









New contributor




Sobuch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct. 












error: use of deleted function 'A::A(const A&)'
 return tmp;
 ^~~


Why is the copy constructor called only when there is a virtual destructor in A? How to avoid this?



struct B ;

struct A
 std::unique_ptr<B> x;
 virtual ~A() = default;
;

A f() 
 A tmp;
 return tmp;






c++




c++






share|improve this question









New contributor




Sobuch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Sobuch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 3 hours ago







Sobuch













New contributor




Sobuch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









SobuchSobuch

485




485




New contributor




Sobuch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Sobuch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Sobuch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • see: In which situations is the C++ copy constructor called?

    – kmdreko
    3 hours ago






  • 3





    C++ handles objects different than C#/Java. When an instance goes out of scope (tmp here) its destructor must be called. Therefore, when you return tmp then you're asking it to make a copy of tmp to be return to whomever calls the function. Once copied, tmp will be destroyed and its copy will be available for use.

    – Everyone
    3 hours ago

















  • see: In which situations is the C++ copy constructor called?

    – kmdreko
    3 hours ago






  • 3





    C++ handles objects different than C#/Java. When an instance goes out of scope (tmp here) its destructor must be called. Therefore, when you return tmp then you're asking it to make a copy of tmp to be return to whomever calls the function. Once copied, tmp will be destroyed and its copy will be available for use.

    – Everyone
    3 hours ago
















see: In which situations is the C++ copy constructor called?

– kmdreko
3 hours ago





see: In which situations is the C++ copy constructor called?

– kmdreko
3 hours ago




3




3





C++ handles objects different than C#/Java. When an instance goes out of scope (tmp here) its destructor must be called. Therefore, when you return tmp then you're asking it to make a copy of tmp to be return to whomever calls the function. Once copied, tmp will be destroyed and its copy will be available for use.

– Everyone
3 hours ago





C++ handles objects different than C#/Java. When an instance goes out of scope (tmp here) its destructor must be called. Therefore, when you return tmp then you're asking it to make a copy of tmp to be return to whomever calls the function. Once copied, tmp will be destroyed and its copy will be available for use.

– Everyone
3 hours ago












1 Answer
1






active

oldest

votes


















20














virtual ~A() = default; is a user declared destructor. Because of that, A no longer has a move constructor. That means return tmp; can't move tmp and since tmp is not copyable, you get a compiler error.



There are two ways you can fix this. You can add a move constructor like



struct A
 std::unique_ptr<B> x;

 A() = default; // you have to add this since the move constructor was added
 A(A&&) = default; // defaulted move
 virtual ~A() = default;
;


or you can create a base class that has the virtual destructor and inherit from that like



struct C 
 virtual ~C() = default;
;

struct A : C 
 std::unique_ptr<B> x;
;





share|improve this answer




















  • 6





    Better to follow the Rule of Zero/Five. Either add all of (copy ctor, move ctor, copy assignment, move assignment, destructor) or add none of them. In this example, none of them are necessary.

    – 0x5453
    3 hours ago






  • 2





    @0x5453 Unless this is a parent class and the OP wants the derived classes to get destroyed properly. You need a virtual destructor if you have polymorphism.

    – NathanOliver
    3 hours ago






  • 2





    @Tzalumen no delete is required (because that's what the unique pointer does for you), but a virtual destructor is required so that the unique pointer won't have UB.

    – eerorika
    3 hours ago







  • 3





    @Tzalumen If you delete an object of class X through a pointer to a base class of X and that base class doesn't have a virtual dtor, it's Undefined Behaviour. Regardless of what the destructor does.

    – Angew
    3 hours ago






  • 3





    @Tzalumen If you have polymorphism, you must have a virtual destructor. If you don't the destructor for the derived class won't be called and you have UB.

    – NathanOliver
    3 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









20














virtual ~A() = default; is a user declared destructor. Because of that, A no longer has a move constructor. That means return tmp; can't move tmp and since tmp is not copyable, you get a compiler error.



There are two ways you can fix this. You can add a move constructor like



struct A
 std::unique_ptr<B> x;

 A() = default; // you have to add this since the move constructor was added
 A(A&&) = default; // defaulted move
 virtual ~A() = default;
;


or you can create a base class that has the virtual destructor and inherit from that like



struct C 
 virtual ~C() = default;
;

struct A : C 
 std::unique_ptr<B> x;
;





share|improve this answer




















  • 6





    Better to follow the Rule of Zero/Five. Either add all of (copy ctor, move ctor, copy assignment, move assignment, destructor) or add none of them. In this example, none of them are necessary.

    – 0x5453
    3 hours ago






  • 2





    @0x5453 Unless this is a parent class and the OP wants the derived classes to get destroyed properly. You need a virtual destructor if you have polymorphism.

    – NathanOliver
    3 hours ago






  • 2





    @Tzalumen no delete is required (because that's what the unique pointer does for you), but a virtual destructor is required so that the unique pointer won't have UB.

    – eerorika
    3 hours ago







  • 3





    @Tzalumen If you delete an object of class X through a pointer to a base class of X and that base class doesn't have a virtual dtor, it's Undefined Behaviour. Regardless of what the destructor does.

    – Angew
    3 hours ago






  • 3





    @Tzalumen If you have polymorphism, you must have a virtual destructor. If you don't the destructor for the derived class won't be called and you have UB.

    – NathanOliver
    3 hours ago
















20














virtual ~A() = default; is a user declared destructor. Because of that, A no longer has a move constructor. That means return tmp; can't move tmp and since tmp is not copyable, you get a compiler error.



There are two ways you can fix this. You can add a move constructor like



struct A
 std::unique_ptr<B> x;

 A() = default; // you have to add this since the move constructor was added
 A(A&&) = default; // defaulted move
 virtual ~A() = default;
;


or you can create a base class that has the virtual destructor and inherit from that like



struct C 
 virtual ~C() = default;
;

struct A : C 
 std::unique_ptr<B> x;
;





share|improve this answer




















  • 6





    Better to follow the Rule of Zero/Five. Either add all of (copy ctor, move ctor, copy assignment, move assignment, destructor) or add none of them. In this example, none of them are necessary.

    – 0x5453
    3 hours ago






  • 2





    @0x5453 Unless this is a parent class and the OP wants the derived classes to get destroyed properly. You need a virtual destructor if you have polymorphism.

    – NathanOliver
    3 hours ago






  • 2





    @Tzalumen no delete is required (because that's what the unique pointer does for you), but a virtual destructor is required so that the unique pointer won't have UB.

    – eerorika
    3 hours ago







  • 3





    @Tzalumen If you delete an object of class X through a pointer to a base class of X and that base class doesn't have a virtual dtor, it's Undefined Behaviour. Regardless of what the destructor does.

    – Angew
    3 hours ago






  • 3





    @Tzalumen If you have polymorphism, you must have a virtual destructor. If you don't the destructor for the derived class won't be called and you have UB.

    – NathanOliver
    3 hours ago














20












20








20







virtual ~A() = default; is a user declared destructor. Because of that, A no longer has a move constructor. That means return tmp; can't move tmp and since tmp is not copyable, you get a compiler error.



There are two ways you can fix this. You can add a move constructor like



struct A
 std::unique_ptr<B> x;

 A() = default; // you have to add this since the move constructor was added
 A(A&&) = default; // defaulted move
 virtual ~A() = default;
;


or you can create a base class that has the virtual destructor and inherit from that like



struct C 
 virtual ~C() = default;
;

struct A : C 
 std::unique_ptr<B> x;
;





share|improve this answer















virtual ~A() = default; is a user declared destructor. Because of that, A no longer has a move constructor. That means return tmp; can't move tmp and since tmp is not copyable, you get a compiler error.



There are two ways you can fix this. You can add a move constructor like



struct A
 std::unique_ptr<B> x;

 A() = default; // you have to add this since the move constructor was added
 A(A&&) = default; // defaulted move
 virtual ~A() = default;
;


or you can create a base class that has the virtual destructor and inherit from that like



struct C 
 virtual ~C() = default;
;

struct A : C 
 std::unique_ptr<B> x;
;






share|improve this answer














share|improve this answer



share|improve this answer








edited 3 hours ago

























answered 3 hours ago









NathanOliverNathanOliver

96k16136209




96k16136209







  • 6





    Better to follow the Rule of Zero/Five. Either add all of (copy ctor, move ctor, copy assignment, move assignment, destructor) or add none of them. In this example, none of them are necessary.

    – 0x5453
    3 hours ago






  • 2





    @0x5453 Unless this is a parent class and the OP wants the derived classes to get destroyed properly. You need a virtual destructor if you have polymorphism.

    – NathanOliver
    3 hours ago






  • 2





    @Tzalumen no delete is required (because that's what the unique pointer does for you), but a virtual destructor is required so that the unique pointer won't have UB.

    – eerorika
    3 hours ago







  • 3





    @Tzalumen If you delete an object of class X through a pointer to a base class of X and that base class doesn't have a virtual dtor, it's Undefined Behaviour. Regardless of what the destructor does.

    – Angew
    3 hours ago






  • 3





    @Tzalumen If you have polymorphism, you must have a virtual destructor. If you don't the destructor for the derived class won't be called and you have UB.

    – NathanOliver
    3 hours ago













  • 6





    Better to follow the Rule of Zero/Five. Either add all of (copy ctor, move ctor, copy assignment, move assignment, destructor) or add none of them. In this example, none of them are necessary.

    – 0x5453
    3 hours ago






  • 2





    @0x5453 Unless this is a parent class and the OP wants the derived classes to get destroyed properly. You need a virtual destructor if you have polymorphism.

    – NathanOliver
    3 hours ago






  • 2





    @Tzalumen no delete is required (because that's what the unique pointer does for you), but a virtual destructor is required so that the unique pointer won't have UB.

    – eerorika
    3 hours ago







  • 3





    @Tzalumen If you delete an object of class X through a pointer to a base class of X and that base class doesn't have a virtual dtor, it's Undefined Behaviour. Regardless of what the destructor does.

    – Angew
    3 hours ago






  • 3





    @Tzalumen If you have polymorphism, you must have a virtual destructor. If you don't the destructor for the derived class won't be called and you have UB.

    – NathanOliver
    3 hours ago








6




6





Better to follow the Rule of Zero/Five. Either add all of (copy ctor, move ctor, copy assignment, move assignment, destructor) or add none of them. In this example, none of them are necessary.

– 0x5453
3 hours ago





Better to follow the Rule of Zero/Five. Either add all of (copy ctor, move ctor, copy assignment, move assignment, destructor) or add none of them. In this example, none of them are necessary.

– 0x5453
3 hours ago




2




2





@0x5453 Unless this is a parent class and the OP wants the derived classes to get destroyed properly. You need a virtual destructor if you have polymorphism.

– NathanOliver
3 hours ago





@0x5453 Unless this is a parent class and the OP wants the derived classes to get destroyed properly. You need a virtual destructor if you have polymorphism.

– NathanOliver
3 hours ago




2




2





@Tzalumen no delete is required (because that's what the unique pointer does for you), but a virtual destructor is required so that the unique pointer won't have UB.

– eerorika
3 hours ago






@Tzalumen no delete is required (because that's what the unique pointer does for you), but a virtual destructor is required so that the unique pointer won't have UB.

– eerorika
3 hours ago





3




3





@Tzalumen If you delete an object of class X through a pointer to a base class of X and that base class doesn't have a virtual dtor, it's Undefined Behaviour. Regardless of what the destructor does.

– Angew
3 hours ago





@Tzalumen If you delete an object of class X through a pointer to a base class of X and that base class doesn't have a virtual dtor, it's Undefined Behaviour. Regardless of what the destructor does.

– Angew
3 hours ago




3




3





@Tzalumen If you have polymorphism, you must have a virtual destructor. If you don't the destructor for the derived class won't be called and you have UB.

– NathanOliver
3 hours ago






@Tzalumen If you have polymorphism, you must have a virtual destructor. If you don't the destructor for the derived class won't be called and you have UB.

– NathanOliver
3 hours ago













Sobuch is a new contributor. Be nice, and check out our Code of Conduct.









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